Bài 6: (0,75 điểm): Cho A = $\frac{1}{2^2}$+ $\frac{1}{3^2}$+ $\frac{1}{4^2}$+ $\frac{1}{5^2}$+...+ $\frac{1}{2021^2}$  Chứng minh rằng A<1.

Đặt $A=\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+...+\dfrac{1}{2021^{2}}$
Ta có: 
$\dfrac{1}{2^{2}}<\dfrac{1}{1.2}$
$\dfrac{1}{3^{2}}<\dfrac{1}{2.3}$
$\dfrac{1}{4^{2}}<\dfrac{1}{3.4}$
...
$\dfrac{1}{2021^{2}}<\dfrac{1}{2020.2021}$
$\Rightarrow A<\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020.2021}$
$\Leftrightarrow A<1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}$
$\Leftrightarrow A<1-\dfrac{1}{2021}$
$\Leftrightarrow A<\dfrac{2020}{2021}$
mà $\dfrac{2020}{2021}<1$
$\Rightarrow A<1$